5. Consider the following information: 1st ionization energy of Na(g) = 495.8 kJ/mol Bond dissociation energy of O2(g) = 498.4 kJ/mol 1st electron affinity of O(g)=-142.5 kJ/mol 2nd electron affinity of O¹(g) = 844 kJ/mol Lattice energy of Na2O(s) = -2608 kJ/mol Enthalpy of formation of Na2O(s) = -416 kJ/mol a Draw the Born-Haber cycle for Na₂O(s). b Calculate the unknown. 120 C Draw the Lewis symbol for Na₂O.

5. Consider the following information: 1st ionization energy of Na(g) = 495.8 kJ/mol Bond dissociation energy of O2(g) = 498.4 kJ/mol 1st electron affinity of O(g)=-142.5 kJ/mol 2nd electron affinity of O¹(g) = 844 kJ/mol Lattice energy of Na2O(s) = -2608 kJ/mol Enthalpy of formation of Na2O(s) = -416 kJ/mol a Draw the Born-Haber cycle for Na₂O(s). b Calculate the unknown. 120 C Draw the Lewis symbol for Na₂O.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter5: Electron Configurations And The Periodic Table

Section: Chapter Questions

Problem 96QRT

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The lattice energy of an ionic solid such as NaCl is the enthalpy change H for the process in which the solid changes to ions. For example, NaCl(s)Na+(g)+Cl(g)H=786kJ/mol Assume that the ionization energy and electron a affinity are H values for the processes defined by those terms. The ionization energy of Na is 496 kJ/mol. Use this, the electron affinity from Table 8.4, and the lattice energy of NaCl to calculate H for the following process: Na(g)+Cl(g)NaCl(s)
Determine the lattice energy for LiCl(s) given these data: Sublimation enthalpy of Li, 161 kJ/mol; IE, for Li, 520 kJ/mol; BE of Cl2(g), 242 kJ/mol; electron affinity of Cl, 349 kJ/mol; formation enthalpy of LiCl(s), 408.7 kJ/mol.
Write the electron configuration for each of the following ions: (a) As3 (b) I (c) Be2+ (d) Cd2+ (e) O2 (f) Ga3+ (g) Li+ (h) N3 (j) Sn2+ (j) Co2+ (k) Fe2+ (1) As3+
Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than thelattice energy of NaCl? Explain.Electron affinity of Cl = –349 kJ/mol1st ionization energy of Rb = 403 kJ/molBond energy of Cl2 = 242 kJ/molSublimation energy of Rb = 86.5 kJ/molΔHf [RbCl (s)] = –430.5 kJ/mol
Given the following information, construct a Born-Haber cycle to calculate the lattice energy of CrCl₂I(s): Net energy change for the formation of CrCl₂I(s) = -420 kJ/mol Bond dissociation energy for I2(g) = +243 kJ/mol Bond dissociation energy for Cl2(g) for Cl2(g) for 12(g) = +151 kJ/mol Heat of sublimation for I2(s) = +62 kJ/mol Heat of sublimation for Cr(s) = +397 kJ/mol = E₁₁ for Cr(g) = 652 kJ/mol E₁₂ for Cr(g) == 1588 kJ/mol E₁3 for Cr(g) = 2882 kJ/mol Eea for Cl(g)=-349 kJ/mol === Eea for I(g) = -295 kJ/mol
Consider a hypothetical ionic compound AB (comprised of A* and B ions). Given the following enthalpy data and using a Born-Haber cycle calculation, predict AHiattice in kJ mol. AG) + Bs) - AB 4H = -384 kJ mol As) → Ag) ArH = 105 kJ mol B(s) - Bg) A,H = 101 kJ mol1 First ionization energy of Ag) = 501 kJ mol" Answes=? Electron affinity enthalpy of Bro (exothermic) = -348 kJ mol
1) Calculate the lattice energy for NaCl(s) using a Born-Haber cycle and the following information: NaCl(s) → Nat(g) + Cl-(g) Na(s) + 1/2 C12(g) → NaCl(s) Na(s) → Na(g) Na(g) → Na+(g) + e- 1/2 C12(g) → Cl(g) Cl(g) + e- → Cl-(g) ? -411.0 kJ/mol +107.3 kJ/mol +495.8 kJ/mol +121.7 kJ/mol -348.6 kJ/mol
Estimate the change in enthalpy (delta H) for the dissociation for HCl(g) into H+(g) and Cl-(g) Ionization energy for H 1312 kJ/mol Electron affinity for H (-72.8 kJ/mol) Ionization energy for Cl 1251.2 kJ/mol Electron affinity for Cl (-349 kJ/mol)
Calculate the lattice energy for LiBr(s) given the following: sublimation energy for Li(s) +166 kJ/mol ΔHf for Br(g) +97 kJ/mol first ionization energy of Li(g) +520. kJ/mol electron affinity of Br(g) –325 kJ/mol enthalpy of formation of LiBr(s) –351 kJ/mol
8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these
Calculate the lattice energy of NaBr(s), given the following thermochemical equations, where A/E and AEA are ionization energy and electron affinity, respectively. Na(s)Na(g) AH = +107 kJ Na(g) Nat(g) + e A/E = +496 kJ -> 1/2 Br₂(g) → Br(g) AHf = +112 kJ - Br(g) + e¯ → Br¯(g) AEA = -325 kJ Na(s) + 1/2 Br₂(g) → NaBr(s) AH = -361 kJ ->> - -1401 kJ -751 kJ +29 kJ -29 kJ +751 kJ
Given the following information: Li(s) → Li(g) Li(s) = 161 kJ/mol Heat of sublimation of HCI(g) → H(g) + Cl(g) Bond energy of HCI = 427 kJ/mol Li(g) → Li+(g) + e- lonization energy of Li(g) = 520. kJ/mol CI(g) + e- → CI-(g) Electron affinity of Cl(g) = -349 kJ/mol Li+(g) + Cl-(g) →→ LiC(s) Lattice energy of LİCI(s) = -829 kJ/mol H2(g) → 2H(g) Bond energy of H2 = 432 kJ/mol calculate the net change in energy for the reaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g) -179 kJ 362 kJ -70 kJ -572 kJ None ese

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